Added 2017
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2017/01/1.md
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2017/01/1.md
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## \-\-- Day 1: Inverse Captcha \-\--
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The night before Christmas, one of Santa\'s Elves calls you in a panic.
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\"The printer\'s broken! We can\'t print the *Naughty or Nice List*!\"
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By the time you make it to [sub-basement
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17]{title="Floor 17: cafeteria, printing department, and experimental organic digitization equipment."},
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there are only a few minutes until midnight. \"We have a big problem,\"
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she says; \"there must be almost *fifty* bugs in this system, but
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nothing else can print The List. Stand in this square, quick! There\'s
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no time to explain; if you can convince them to pay you in *stars*,
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you\'ll be able to\--\" She pulls a lever and the world goes blurry.
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When your eyes can focus again, everything seems a lot more pixelated
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than before. She must have sent you inside the computer! You check the
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system clock: *25 milliseconds* until midnight. With that much time, you
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should be able to collect all *fifty stars* by December 25th.
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Collect stars by solving puzzles. Two puzzles will be made available on
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each ~~day~~ millisecond in the Advent calendar; the second puzzle is
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unlocked when you complete the first. Each puzzle grants *one star*.
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Good luck!
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You\'re standing in a room with \"digitization quarantine\" written in
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LEDs along one wall. The only door is locked, but it includes a small
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interface. \"Restricted Area - Strictly No Digitized Users Allowed.\"
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It goes on to explain that you may only leave by solving a
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[captcha](https://en.wikipedia.org/wiki/CAPTCHA) to prove you\'re *not*
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a human. Apparently, you only get one millisecond to solve the captcha:
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too fast for a normal human, but it feels like hours to you.
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The captcha requires you to review a sequence of digits (your puzzle
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input) and find the *sum* of all digits that match the *next* digit in
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the list. The list is circular, so the digit after the last digit is the
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*first* digit in the list.
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For example:
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- `1122` produces a sum of `3` (`1` + `2`) because the first digit
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(`1`) matches the second digit and the third digit (`2`) matches the
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fourth digit.
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- `1111` produces `4` because each digit (all `1`) matches the next.
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- `1234` produces `0` because no digit matches the next.
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- `91212129` produces `9` because the only digit that matches the next
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one is the last digit, `9`.
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*What is the solution* to your captcha?
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Your puzzle answer was `1158`.
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## \-\-- Part Two \-\-- {#part2}
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You notice a progress bar that jumps to 50% completion. Apparently, the
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door isn\'t yet satisfied, but it did emit a *star* as encouragement.
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The instructions change:
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Now, instead of considering the *next* digit, it wants you to consider
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the digit *halfway around* the circular list. That is, if your list
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contains `10` items, only include a digit in your sum if the digit
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`10/2 = 5` steps forward matches it. Fortunately, your list has an even
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number of elements.
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For example:
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- `1212` produces `6`: the list contains `4` items, and all four
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digits match the digit `2` items ahead.
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- `1221` produces `0`, because every comparison is between a `1` and a
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`2`.
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- `123425` produces `4`, because both `2`s match each other, but no
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other digit has a match.
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- `123123` produces `12`.
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- `12131415` produces `4`.
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*What is the solution* to your new captcha?
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Your puzzle answer was `1132`.
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Both parts of this puzzle are complete! They provide two gold stars:
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\*\*
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At this point, you should [return to your Advent calendar](/2017) and
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try another puzzle.
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If you still want to see it, you can [get your puzzle
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input](1/input).
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2017/01/solution.py
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#!/bin/python3
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import sys
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from pprint import pprint
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input_f = "input" #sys.argv[1]
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arr = []
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with open(input_f) as file:
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for line in file:
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arr = list(line.rstrip('\n'))
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#########################################
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# #
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# Part 1 #
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# #
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#########################################
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sum = 0
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for idx,x in enumerate(arr):
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if idx != len(arr)-1:
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if x == arr[idx+1]:
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sum+=int(x)
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else:
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if x == arr[0]:
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sum+=int(x)
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print(sum)
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#########################################
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# #
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# Part 2 #
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# #
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#########################################
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sum=0
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length=len(arr)
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for idx,x in enumerate(arr):
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if idx != length:
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print(x,arr[(int(length/2)+idx) % length])
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if x == arr[(int(length/2)+idx) % length]:
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sum+=int(x)
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print(sum)
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2017/02/2.md
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## \-\-- Day 2: Corruption Checksum \-\--
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As you walk through the door, a glowing humanoid shape yells in your
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direction. \"You there! Your state appears to be idle. Come help us
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repair the corruption in this spreadsheet - if we take another
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millisecond, we\'ll have to display an hourglass cursor!\"
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The spreadsheet consists of rows of apparently-random numbers. To make
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sure the recovery process is on the right track, they need you to
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calculate the spreadsheet\'s *checksum*. For each row, determine the
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difference between the largest value and the smallest value; the
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checksum is the sum of all of these differences.
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For example, given the following spreadsheet:
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5 1 9 5
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7 5 3
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2 4 6 8
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- The first row\'s largest and smallest values are `9` and `1`, and
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their difference is `8`.
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- The second row\'s largest and smallest values are `7` and `3`, and
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their difference is `4`.
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- The third row\'s difference is `6`.
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In this example, the spreadsheet\'s checksum would be `8 + 4 + 6 = 18`.
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*What is the checksum* for the spreadsheet in your puzzle input?
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Your puzzle answer was `34925`.
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The first half of this puzzle is complete! It provides one gold star: \*
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## \-\-- Part Two \-\-- {#part2}
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\"Great work; looks like we\'re on the right track after all. Here\'s a
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*star* for your effort.\" However, the program seems a little worried.
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Can programs *be* worried?
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\"Based on what we\'re seeing, it looks like all the User wanted is some
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information about the *evenly divisible values* in the spreadsheet.
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Unfortunately, none of us are equipped for that kind of calculation -
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most of us specialize in [bitwise
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operations]{title="Bonus points if you solve this part using only bitwise operations."}.\"
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It sounds like the goal is to find the only two numbers in each row
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where one evenly divides the other - that is, where the result of the
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division operation is a whole number. They would like you to find those
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numbers on each line, divide them, and add up each line\'s result.
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For example, given the following spreadsheet:
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5 9 2 8
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9 4 7 3
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3 8 6 5
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- In the first row, the only two numbers that evenly divide are `8`
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and `2`; the result of this division is `4`.
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- In the second row, the two numbers are `9` and `3`; the result is
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`3`.
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- In the third row, the result is `2`.
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In this example, the sum of the results would be `4 + 3 + 2 = 9`.
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What is the *sum of each row\'s result* in your puzzle input?
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Answer:
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Although it hasn\'t changed, you can still [get your puzzle
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input](2/input).
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2017/02/solution.py
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#!/bin/python3
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import sys
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from pprint import pprint
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input_f = "input" #sys.argv[1]
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sum = 0
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with open(input_f) as file:
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for line in file:
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line = list(map(int,line.rstrip('\n').split()))
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#########################################
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# #
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# Part 1 #
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# #
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#########################################
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sum+=(max(line)-min(line))
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print(sum)
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#########################################
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# #
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# Part 2 #
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# #
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#########################################
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